Question

A capacitor of capacitance 2$\mu$F if connected as shown in figure. The internal resistance of the cell is 0.5 $\Omega$. The amount of charge on the capacitor plates is :

• A. Zero
• B. 2 $\mu C$
• C. 4 $\mu C$
• D. 6 $\mu C$

Answer 1

The correct option is C 4 $\mu C$

Under steady state, current through 10 $\Omega$ resistor is zero.
Current through battery is, i = $\frac{2.5}{2+0.5}=1A$
From KVL, 2.5 - 0.5 $\times 1=V_c$
$\therefore V_c=2V$
Change on capacitor, q = CV = $2\mu \times 2=4\mu C$